Eigenvalues and eigenvectors

A nonzero vector $\mathbf v$ is an eigenvector of a square matrix $A$ ($n\times n$) with eigenvalue $\lambda$ if multiplying by $A$ reproduces $\mathbf v$ scaled by $\lambda$:

where: $A$ is the $n\times n$ matrix, $\mathbf v$ is the eigenvector (a nonzero column vector), and $\lambda$ is the eigenvalue (a scalar, possibly complex or repeated). Rearranging gives $(A-\lambda I)\mathbf v=\mathbf 0$. A homogeneous system has a nonzero solution only when its coefficient matrix is singular, so eigenvalues are exactly the roots of the characteristic equation

a degree-$n$ polynomial in $\lambda$ whose $n$ roots (with multiplicity) are the eigenvalues of $A$. For each root $\lambda$, the eigenvectors are the nonzero solutions of $(A-\lambda I)\mathbf v=\mathbf 0$; together with $\mathbf 0$ they span the eigenspace $E_\lambda=\ker(A-\lambda I)$. Every eigenvalue carries two multiplicities that need not agree: its algebraic multiplicity is how many times it repeats as a root of the characteristic polynomial, while its geometric multiplicity is $\dim E_\lambda$, the number of independent eigenvectors it actually supplies. Always $1\le$ geometric $\le$ algebraic multiplicity. When every eigenvalue has geometric multiplicity equal to its algebraic multiplicity — guaranteed whenever all $n$ eigenvalues are distinct, since eigenvectors for distinct eigenvalues are automatically linearly independent — $A$ has a full set of $n$ independent eigenvectors and is diagonalizable. Two quick facts turn hand computation into something self-checking rather than a leap of faith: the sum of the eigenvalues equals $\operatorname{tr}(A)$, and their product equals $\det A$. A third fact needs no computation at all — for a triangular (or diagonal) matrix, the eigenvalues are simply the diagonal entries, read off directly. A few more facts are worth carrying forward from the earlier topic on special matrices, cited here without re-derivation: eigenvalues of a Hermitian or symmetric matrix are always real, eigenvalues of a unitary or orthogonal matrix always have modulus $1$, and eigenvectors of a symmetric or Hermitian matrix corresponding to distinct eigenvalues are automatically orthogonal to each other. It is worth being explicit about why an eigenspace is a genuine subspace rather than just a convenient set: $E_\lambda=\ker(A-\lambda I)$ is the null space of a linear map, so it automatically contains $\mathbf 0$ and is closed under addition and scalar multiplication — any two eigenvectors sharing the same $\lambda$ can be added or scaled and the result is again either an eigenvector for that $\lambda$ or the zero vector. This is what lets you speak of a basis for $E_\lambda$ and hence of its dimension, the geometric multiplicity, rather than just listing individual eigenvectors one at a time.

Problem. Find the eigenvalues and eigenvectors of $A=\begin{pmatrix}4&1\\2&3\end{pmatrix}$, and check whether $A$ is diagonalizable.

Step 1. Set up the characteristic equation: $\det(A-\lambda I)=\det\begin{pmatrix}4-\lambda&1\\2&3-\lambda\end{pmatrix}=(4-\lambda)(3-\lambda)-2=0$.

Step 2. Expand and solve: $12-7\lambda+\lambda^2-2=0\Rightarrow \lambda^2-7\lambda+10=0\Rightarrow(\lambda-5)(\lambda-2)=0$, giving $\lambda_1=5$, $\lambda_2=2$. Check: $\lambda_1+\lambda_2=7=\operatorname{tr}(A)=4+3$ and $\lambda_1\lambda_2=10=\det A=12-2$ — both match, so the roots are trustworthy.

Step 3. Find the eigenvector for $\lambda_1=5$ by solving $(A-5I)\mathbf v=\mathbf 0$: $\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}=\mathbf 0$ gives $-v_1+v_2=0$, so $v_1=v_2$ and $\mathbf v_1=(1,1)$.

Step 4. Find the eigenvector for $\lambda_2=2$ by solving $(A-2I)\mathbf v=\mathbf 0$: $\begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}=\mathbf 0$ gives $2v_1+v_2=0$, so $\mathbf v_2=(1,-2)$. Both eigenvalues are distinct (each algebraic and geometric multiplicity 1), and $\{(1,1),(1,-2)\}$ are linearly independent, so $A$ has 2 independent eigenvectors and is diagonalizable.

Competitive stretch

For $A=\begin{pmatrix}2&1&0\\0&2&1\\0&0&3\end{pmatrix}$ (upper triangular), the eigenvalues are the diagonal entries directly: $\lambda=2$ with algebraic multiplicity $2$, and $\lambda=3$ with algebraic multiplicity $1$. Solving $(A-2I)\mathbf v=\mathbf 0$ gives $v_2=0,v_3=0$ with $v_1$ free, so $E_2=\operatorname{span}\{(1,0,0)\}$ — only one independent eigenvector, geometric multiplicity $1<2$. Since geometric multiplicity falls short of algebraic multiplicity for $\lambda=2$, this $3\times3$ matrix has only $2$ independent eigenvectors total and is not diagonalizable, even though its eigenvalues were trivial to read off.

• Solving $\det(A-\lambda I)=0$ correctly, then plugging $\lambda$ back into the wrong matrix — always substitute into $A-\lambda I$ (not $A$) when finding each eigenvector, re-solving the homogeneous system separately for every $\lambda$.
• Assuming geometric multiplicity always equals algebraic multiplicity — the competitive stretch example shows a repeated eigenvalue can have a strictly smaller eigenspace, which is exactly what blocks diagonalizability.
• Forgetting $\mathbf v\neq\mathbf 0$ is part of the definition — the zero vector trivially satisfies $A\mathbf 0=\lambda\mathbf 0$ for every $\lambda$ and is never an eigenvector.
• Sign or arithmetic slips expanding a $3\times3$ determinant — always cross-check with $\operatorname{tr}(A)=\sum\lambda_i$ and $\det A=\prod\lambda_i$.

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